Find the value of t such that the set $A={ (x_1, x_2, x_3, x_4) | 3x_1-2x_2+x_3+x_4-t(x_1^2+x_4^2)=0}$ be a subspace of $\mathbb R^4$

Question

Find the value of $t\in \mathbb{R}$ such that the set $A= \{ (x_1, x_2, x_3, x_4) | 3x_1-2x_2+x_3+x_4-t(x_1^2+x_4^2)=0\} $ is a subspace of $\mathbb R^4$.

I know that for a set to be a subspace of $\mathbb R^4$ should have the vector zero but I don't see how can that help me find the value $t$. I would really appreciate any help.

Answer 1

Obviously $0\in A$. According to definition if $x=(x_1,x_2,x_3,x_4)\in A$ and $y=(y_1,y_2,y_3,y_4)\in A$ we must have $ax+by\in A$ for any $a,b\in\Bbb R$ i.e.$$3(ax_1+by_1)-2(ax_2+by_2)+ax_3+by_3+ax_4+by_4-t\left((ax_1+by_1)^2+(ax_4+by_4)^2\right)=0$$Let $b=0$. From the other side according to definition$$3x_1-2x_2+x_3+x_4-t(x_1^2+x_4^2)=0$$which finally yields $$t(a-a^2)(x_1^2+x_4^2)=0$$which must be true for all $a$ and $x_1$ and $x_4$ therefore$$t=0$$

Answer 2

For a subspace we need $$x=(x_1,x_2,x_3,x_4)\in A \implies \lambda x \in A$$

Now if $(x_1,x_2,x_3,x_4)\in A$ we have

$$ 3x_1-2x_2+x_3+x_4-t(x_1^2+x_4^2)=0 $$ and $$\lambda (3x_1-2x_2+x_3+x_4)-\lambda^2t(x_1^2+x_4^2)=0 $$

Thus for every $\lambda$ you have $$(\lambda^2-\lambda)t(x_1^2+x_4^2)=0$$

Which implies $t=0$.